Which of the following is the correctly balanced chemical equation, including phases, for the reaction of aqueous Pb(ClO3)2 with aqueous NaI?

• A. Pb(ClO3)2(aq) + 2 NaI(aq) → PbI2(aq) + 2 NaClO3(s)
• B. Pb(ClO3)2(aq) + NaI(aq) → PbI2(aq) + NaClO3(s)
• C. Pb(ClO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaClO3(aq)
• D. Pb(ClO3)2(aq) + 2 NaI(aq) → PbI2(s) + NaClO3(aq)

Answer: (C)

Verified step by step guidance

1. Identify the reactants and products along with their phases: Pb(ClO3)2 and NaI are both aqueous (aq), and the products are PbI2 and NaClO3 with their respective phases to be determined based on solubility rules.
2. Apply the solubility rules to determine the phases of the products: PbI2 is generally insoluble in water and will precipitate as a solid (s), while NaClO3 is soluble and remains aqueous (aq).
3. Write the unbalanced chemical equation with correct formulas and phases: \(\mathrm{Pb(ClO_3)_2(aq) + NaI(aq) \rightarrow PbI_2(?) + NaClO_3(?)}\).
4. Balance the chemical equation by ensuring the number of atoms of each element is the same on both sides: balance iodine (I) atoms by placing a coefficient of 2 before NaI and PbI2, and balance chlorate (ClO3) groups accordingly.
5. Confirm the final balanced equation with correct phases: \(\mathrm{Pb(ClO_3)_2(aq) + 2\ NaI(aq) \rightarrow PbI_2(s) + 2\ NaClO_3(aq)}\).